To find #E_(cell)# when we do not have standard conditions we use the **Nernst Equation:**

#E_(cell)=E_(cell)^@-(RT)/(nF)lnQ#

At 298K this simplifies to:

#E_(cell)=E_(cell)^@-(0.05916)/(n)logQ#

#n# is the number of moles of electrons transferred.

#Q# is the reaction quotient.

To find #E_(cell)^@# and the cell reaction list the #E^@# values most -ve to most +ve :

#" "E^@"(V)"#

#stackrelcolor(blue)(larr)color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx)#

#Sn^(2+)+2erightleftharpoonsSn" "-0.136#

#TiO^(2+)+e+2H^+rightleftharpoonsTi^(3+)+H_2O" "+0.1#

#stackrelcolor(red)(rarr)color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx)#

The most +ve 1/2 cell is the one that takes in the electrons so you can see that #TiO^(2+)# will oxidise #Sn# to #Sn^(2+)#

The cell reaction is therefore:

#Sn_((s))+2TiO_((aq))^(2+)+4H_((aq))^(+)rarrSn_((aq))^(2+)+2Ti_((aq))^(3+)+2H_2O_((l))#

The reaction quotient is therefore given by:

#Q=([Sn^(2+)][Ti^(3+)]^2)/([TiO^(2+)]^(2)[H^+]^4)#

To find #E_(cell)^@# subtract the **least +ve** 1/2 cell from the **most +ve**#rArr#

#E_(cell)^@=+0.1-(-0.136)=+0.236"V"#

Now insert the values into the Nernst Equation:

#E_(cell)=0.236-(0.05916)/(2)xxlog[((0.1)(0.1)^2)/((0.2)^2(0.05^2)]]#

#E_(cell)=0.236-(0.05916)/(2)log[400]#

#E_(cell)=0.236-[0.02958xx(2.6)]#

#E_(cell)=0.236-0.077#

#E_(cell)=+0.16"V"#